# Probability Question

Here's the question:

In a city there are 3 motorcycle sellers. Each seller make it's own motorcycle and sell it to people. In a day about 1000 motorcycles are sold in the city. The probability that a seller's bike would be sold is given by $$P_n$$ where n is the particular seller. Given $$P_1 = 2/8$$, $$P_2 = 4/8$$, $$P_3 = 6/8$$, find the expected number of motorcycles sold by each seller.

I made this question. I am not sure if the question is technically correct or not. I further doubt if $$P_1 + P_2 + P_3 = 1$$ is a necessary condition or not.

Help me figuring it out.

Note by Lokesh Sharma
4 years, 8 months ago

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It is consistent, but incomplete. $$P_1 + P_2 + P_3 \neq 1$$ is all fine; what we know is that "among the bikes that the first seller sells, it's expected that $$P_1$$ of them are actually sold", and so on. We're missing the number of bikes that each seller plans to sell. However, assuming the numbers are identical, we can solve it:

Suppose each seller makes $$x$$ motorcycles. Then seller $$i$$ is expected to sell $$P_ix$$ motorcycles. In total, there are $$1000$$ motorcycles sold, so $$P_1x + P_2x + P_3x = \frac{3x}{2} = 1000$$, giving $$x = \frac{2000}{3}$$. Thus the expected number of motorcycles sold by the first seller will be $$P_1x = \frac{500}{3}$$, and similarly with other sellers.

- 4 years, 7 months ago

Oh thank you very much! It was easier then I expected; my teacher almost took the entire lecture and couldn't figure it out.

- 4 years, 7 months ago